Solve the following equation:

Sin r =cos(r+20)
With shown working below

4 answers

sin r = cos(r+20)

sin r = cos r cos 20 + sin r sin 20

sin r = cos r cos 20 + sin2 r sin 20

sin r - sin2 r sin 20 = cos r cos 20

sin r (1 - sin 20) = cos r cos 20

sin r = cos r cos 20 / (1 - sin 20)

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20 + 2nπ

where n is an integer.
sin r = cos(r+20)

sin r = cos r cos 20 + sin r sin 20

sin r = cos r cos 20 + sin2 r sin 20

sin r - sin2 r sin 20 = cos r cos 20

sin r (1 - sin 20) = cos r cos 20

sin r = cos r cos 20 / (1 - sin 20)

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20 + 2nπ

where n is an integer.
Solve the following equation:
Sin r =cos(r+20)
With shown working below

asked by Mildred Oshio
just now
2 answers
sin r = cos(r+20)

sin r = cos r cos 20 + sin r sin 20

sin r = cos r cos 20 + sin2 r sin 20

sin r - sin2 r sin 20 = cos r cos 20

sin r (1 - sin 20) = cos r cos 20

sin r = cos r cos 20 / (1 - sin 20)

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20 + 2nπ

where n is an integer.

answered
just now
sin r = cos(r+20)

sin r = cos r cos 20 + sin r sin 20

sin r = cos r cos 20 + sin2 r sin 20

sin r - sin2 r sin 20 = cos r cos 20

sin r (1 - sin 20) = cos r cos 20

sin r = cos r cos 20 / (1 - sin 20)

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20

r = cos-1 (cos r cos 20 / (1 - sin 20)) - 20 + 2nπ

where n is an integer.
AAAaannndd the bot gets it wrong yet again!

Assuming degrees, we have
since cosx = sin(90-x)
r + r+20 = 90
r = 35
check: sin35° = cos55° ✅

there are other solutions in QIII, but judging from the simplicity of the question, I suspect we need only use QI.